A ball rebounds 7 8 as high as it bounced on the previous bounce and is dropped from a height of 8 feet. How high does it bounce on the fourth bounce and how far has it traveled after the fourth bounce?

Question

anonymous · Answer

if thats a (7/8), then a formula that tells you the height of the bounce on the nth bounce is:
$$h = 8(\frac{7}{8})^{n}$$
Thus on the fourth bounce it will be:
$$h = 8(\frac{7}{8})^{4}$$
Unfortunately I dont have a calculator, just type that into one and youre good

anonymous · Answer

I believe the previous poster was incorrect. That is the answer to the FIFTH bounce of the ball (n-1, let's not forget). And as for the unaddressed second portion of the question, just add the sum of all previous calculations, from n=3 to n=0.

anonymous · Answer

i really think that is the equation for the 4th bounce. Its being dropped from 8 feet, and when it bounces the first time it will come back up:
$$8(\frac{7}{8})$$
Notice how that is 7/8 to the first power for the first bounce. So it should be to the 4th power for the fourth bounce.  Now if they had bounced it instead of dropping it, and it reached a height of 8 feet on its first bounce, then my calculation would indeed be for the fifth bounce. 
I dont remember seeing the part about sum, my bad on that part.